Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include<stdio.h>
#include<map>
using namespace std;
typedef long long ll;
const ll mod=1000000000000000031ll,b=1000000000000000003ll,rb=821428571428571454ll;
ll mul(ll a,ll b){return(__int128)a*b%mod;}
ll ad(ll a,ll b){return(a+b)%mod;}
ll pow(ll a,ll b){
	ll s=1;
	while(b){
		if(b&1)s=mul(s,a);
		a=mul(a,a);
		b>>=1;
	}
	return s;
}
char s[250010];
ll p[250010],st[250010];
map<ll,int>m;
int main(){
	int n,i;
	ll ans;
	scanf("%d%s",&n,s+1);
	p[0]=1;
	for(i=1;i<=n;i++){
		p[i]=p[i-1];
		st[i]=st[i-1];
		switch(s[i]){
			case'+':st[i]=ad(st[i],p[i]);break;
			case'-':st[i]=ad(st[i],mod-p[i]);break;
			case'>':p[i]=mul(p[i],b);break;
			case'<':p[i]=mul(p[i],rb);break;
		}
	}
	ans=0;
	for(i=n;i>0;i--){
		m[st[i]]++;
		ans+=m[ad(mul(st[n],p[i-1]),st[i-1])];
	}
	printf("%lld",ans);
}